## Precalculus (6th Edition)

$120$ mL
The given table has the last column with values $\left[\begin{array}{l} 0.05x\\ 0.20(60)=12\\ 0.10(x+60) \end{array}\right]$ The equation to solve is $0.05x+12=0.10(x+60)\qquad$ ... multiply with 100 $5x+1200=10(x+60)$ $5x+1200=10x+600\qquad$ ... add $(-5x-600)$ $600=5x$ $x=\displaystyle \frac{600}{5}=120$ mL